Math 280 Final Part I 6-12-13 Show All Work
1) Find an equation of the plane containing the line
L1 : x  2  t , y  3  t , z  1  t
and a point (2,0,0).
1) To find an equation of a plane, you can first find any three non collinear points in the plane. Since
the line is contained in the plane, all the points on the line is also in the plane. Pick any two
values of t, say t=0 and t=1.
R(2,0,0), we find


t  0  P(2,3,1) , t  1  Q(3,2,2) .

PQ  1,1,1  , PR  0,3,1  .
Together with the given point
Their cross product is <a,b,c>:
PR  4,1,3  . Using (2,0,0) as ( x9 , y0 , z0 ) , we have
4( x  2)  1( y  0)  3( z  0)  0  4 x  y  3z  8

PQ
2) Find an equation of the circle of curvature for
the base point <a,b>, center =
First find the curvature at x=1:
 a, b  

1

y  x 3 at x  1.
(hint: you may use the formula: At
N
f " (1)
3
2 2
(1  f ' (1) )

6
3
2 2
(1  3 )

6
10
3
2
. To find the normal vector, first
parametrize the curve. Then find the unit tangent vector. Finally switch the x and y components of a
(unit) tangent vector and multiply one of the component by -1. By looking at the normal vector, make
sure that the x-component is positive, the y-component is negative.
r(t )  t , t 3  r' (t )  1,3t 2  r' (1)  1,3  .
A unit tangent vector is

1
3
,
.
10 10
The normal
3
vector is
3
1

,
.
10
10
Since the radius is
1


Thus the center is
3
2
5
8
10 2
3
1
 1,1  

,
  1,1    5,  6,  .
3
3
6
10 10
10
5 10
, the equation of the circle of curvature is

6
3
8
250
( x  6) 2  ( y  ) 2 
3
9
3) Find parametric equations for the line of intersection of the planes
2x  y  z  1
and
 2 x  2 y  z  7
(In Math 285, we use the method called row-echelon form to find an equation of the line. In Math 280,
we just find any two solutions to the system of equations, remembering that one of the variables is
free.
First let x=0. Then we have
y=0. We have
 y  z  1
.

 2 y  z  7
2 x  z  1

 2 x  z  7
y  2, z  3 .
P(0,2,-3) is a solution. Next let
. But this system has no solution. Next try z=0. We have
2 x  y  1
3
. We see that Q ( ,2,0) .

2
 2 x  2 y  7
3
x  t , y  2, z  3  3t ,   t  
2
4)
We get

<a,b,c>= PQ

3
,0,3 
2
Using P as
( 5 points)Find the direction of the most rapid decrease of
f ( x, y )  tan 1
The direction of the gradient is the direction of the most rapid increase.
y
1
( 2 )
f f
x ,
x
f  ,


y 2
y 2
x y
1 ( ) 1 ( )
x
x
(1,1)
1 1
  , 
2 2
( x0 , y0 , z0 ) , we have
y
at (1,1).
x
5) For
w  f ( x, y),
x  r 2 s, y  5r  2s , compute
2w
r 2
6) ( 5 points each)
a) Write
0
 
4 x 2
 2 x
f ( x, y)dydx in polar coordinates.
b) Write an integral that represents the volume of the region in the first octant bounded
by y  z  2 and
x 2  y 2  4 using dx dz dy
a) Sketch the region:
3
4
 

2
0
f (r cos  , r sin  )d
2
b) The typical line parallel to the x-axis enter the region through the yz-plane, exits
through the curved surface
x 2  y 2  4 The projection of the region onto the yz-plane is
the triangular region shown below:
V 
2
0

2 y
0

4 y 2
0
1 dxdzdy
Math 280 Final Part II 6-12-13
C shown below for F  x, y 2 



1
1
1
,
,
:
sin

d



1
sin

cos

d



sin 2  cos  d  



4


4
2
6
2
4
4
7) Verify Work Form of Green’s Theorem for the path
(hint:
you may not
need all of the identities)
1) First compute the line integral:
a) First on
line:
c1 , the line segment from (-1,0) to (0,0).
x  t , y  0,  r' (t )  1,0  .  1  t  0 .
b
0
a
1
 F(r(t ))  r' (t )dt  
0
Use common sense to parametrize the
 t ,0  1,0  dt   t dt  
1
b) Next the line segment from (0,0) to
(
1 1
,
).
2 2
1
2
Since the equation of the line is
1
 r ' (t )  1,1  .
2
1
1
b
1
1
2
2  t , t 2  1,1  dt 
F
(
r
(
t
))

r
'
(
t
)
dt

a
0
 02 (t  t )dt  4  6 2
can parametrize the line as
c) Finally the circular path:

b
x  t, y  t, 0  t 
r (t )  cos t , sin t  r ' (t )   sin t , cos t :
 F(r(t ))  r' (t )dt     cos t , sin
a
Thus

C1 C2 C3
4

4
4
t 
1
1
t   sin t , cos t  dt   
4 6 2
1 1
1
1
1
1
  
 

2 4 6 2 4 6 2
2
2) Next using dA:
(
2


(
N M

)dA  
y
x

 1
(2 y  1)dA     (2r sin   1)rdrd
4
0
 2
2r 3
r2
1
2 1
3
sin   ) 10 d    ( sin   )d  (
 1) 
3
2
2
3 2
8
4 3
.
y  x , we
1) Verify Stokes’s theorem for the surface
(hints:

2
0

2
0
sin  d  0 ,
cos 2  d  

2
0
z  x2  y 2 , 0  z  4
cos  d  0 ,

2
0
for
sin  cos  d  0 ,
F  y  z, z  x, x 2 

2
0
sin 2  d   ¸
. You may not need all of the identities)
1) First line integral: the boundary curve is the circle of radius 2: it can be parametrized
(counterclockwise) as
r (t )  2 cos t ,2 sin t ,4 
(note that z=4 since the circle is the intersection
curve of z =4 and the paraboloid.)

F(r(t ))  r' (t )dt  

2
0
2
0
 2 sin t  4,4  2 cos t ,4 cos 2 t  2 sin t ,2 cos t ,0  dt
(4 sin 2 t  8 sin t  8 cos t  4 cos 2 t )dt  4  4  8
 4 sin 2 t  4 cos 2 t  4
(or you can use the identity
and integrate )
2) Next the surface integral:
First compute CurlF:
i

x
yz
j

y
zx
F  y  z, z  x, x 2 
k

 0  1,(2 x  1),1  1  1,2 x  1,2 
z
x2
For the surface equation , move all the variables to one side before taking the gradient:
z  x 2  y 2  x 2  y 2  z  0 . g  2 x,2 y,1 , g  2 x,2 y,1  .
Using counterclockwise
orientation of the curve, the compatible orientation of the surface is upward normal. Thus
n  2 x,2 y,1  .
The surface can be projected onto the xy-plane since every lines perpendicular to the xy-plane
crosses the surface once. The projection is a circle of radius 2.
 g
  curlF   g  p dA   
2
2
0
0
 1,2 x  1,2    2 x,2 y,1 
dA  
 2 x,2 y,1  0,0,1 
 (2 x  4 xy  2 y  2)dA  ( in
  (2r cos  (4(r cos )(r sin  ))  2r sin   2)rdrd . Using the hints, all the
integrals except the last term become 0. Thus   ( 2) dA  2( area )  2( 4 )  8
polar coordinates)
3) ( 5 points each)

a) Write
b)
a)
4
4
0
r
  
2
0
Compute
r dzdrd in spherical coordinates. Do not compute the integral.
3
 
0
First observe that
1
cos x3dxdy
y
3
z  r  z  x2  y2
is a cone
  0 , exits at z  4   
The typical radius enters the region at
V 

2
0

 
4
0
4
cos
0
x
Need to switch dx and dy:
y
 y  3x 2 .
3
4
cos 
. Thus
 2 sin  ddd
3
b)
z  4 is a plane.
3
 
0
1
y
3
 
0
1
y
3
cos x3dxdy
1 3x 2
cos x dxdy  
3

0 0
1
cos x3dydx   3x 2 cos x3dx  sin 1
0
4) Find all the local exterma (maximum, minimum, saddle points) for
f ( x, y ) 
1 3 1 3
x  y  4 xy
3
3
First find all the critical points:
From the first equation,
y
 x( x  64)  0  x  0,4 .
3
are the critical points.
x2
4
f x  x  4 y , f y  y  4x .
2
Solve
 x 2  4 y  0
 2
 y  4 x  0
by substitution:
. Substitute into the second equation, we have
x2 0
x 2 16
x0 y
 0 x4 y

 4.
4
4
4 4
2) Next apply the second derivative test:
At (0,0),
2
Thus (0,0) and (4,4)
f xx ( x, y )  2 x, f yy ( x, y )  2 y, f xy ( x, y )  4
f xx (0,0)  0 , D  f xx (0,0) f yy (0,0)  f xy (0,0) 2  0  16  16 .  (0,0) is a saddle point.
At (4,4) At (0,0),
minimum point.
f xx (4,4)  8 , D  f xx (4,4) f yy (4,4)  f xy (4,4) 2  (8)(8)  16  48  0 .  (4,4) is a local
\
5)
Use Lagrange to find the maximum of
f ( x, y, z )  xz  y with the constraint
g ( x, y, z )  x 2  2 y 2  z 2  6
f x  g x  z    2 x ,
f y  g y  1    4 y ,
f z  g z  x    2 z
There are several approaches in finding x, y, z.
Method1: We can consider cases, depending on the variable is 0 or not.
Case 1)
x  0.
Then by looking at the third equation we can also conclude that
z  0.
In addition,
y  0 since the left side of the second equation is not 0. Thus we can solve for  for each equation.
z
x
1
We have z    2 x   
, 1    4y   
, x    2z   
,
2x
2z
4y
Looking at the first and the third equation,
If
x z,  
z
z 1

 .
2x 2z 2
z
x

 2x2  2z 2  x   z .
2x 2z
Using the second equation,

1
1 1
1
 
y .
4y
2 4y
2
1
11
11
11
g ( x, y, z )  x 2  2 y 2  z 2  6  x 2  2( ) 2  x 2  6  2 x 2   x 2 
x
.
2
2
4
2
11 1 11
11 1
11
, ,
) , (
, ,
)
we get two points: (
2 2 2
2 2
2
If
x  z ,  
z
z
1

 .
2 x 2( z )
2
Using the second equation,

In this case
1
1 1
1
 
y .
4y
2 4y
2
1
11
11
11
g ( x, y, z )  x 2  2 y 2  z 2  6  x 2  2( ) 2  x 2  6  2 x 2   x 2 
x
.
2
2
4
2
11 1 11
11 1
11
, ,
) , (
, ,
)
we get two more points: (
2
2
2
2
2 2
In this case
x  0 . Then by looking at the third equation we conclude that z  0 also. Now go to the
2
2
2
2
2
2
2
constraint, g ( x, y, z )  x  2 y  z  6  0  2 y  0  6  2 y  6  y   3
Case 2)
We get two more points:
(0, 3, 0), (0, 3,0)
We therefore get a total of 6 points that possibly the maximum can be attained..
point
f(x,y,z)
11 1 11
, ,
)
2 2 2
11 1
11
(
, ,
)
2 2
2
11 11 1 13
)(
) 
2
2
2 4
11
11 1 13
(
)( 
) 
2
2
2 4
(
(
11 1
11
, ,
)
2
2
2
11 1
11
(
, ,
)
2
2
2
(0, 3, 0)
11 11 1 9
)(
) 
2
2
2 4
11
11 1 9
(
)( 
) 
2
2
2 4
0 3  3
(0, 3, 0)
0 3   3
(
(
Thus the maximum value is
13
4
Method 2:
We have
f x  g x  z    2 x ,
f y  g y  1    4 y ,
Substitute the first equation into the third:
 x(42  1)  0  x  0,  
1
1
,   .
2
2
f z  g z  x    2 z
z    2x , x    2 z  x   (2( 2 x)  x  42 x .
We get three cases based on these values.
x  0 : Then by looking at the third equation we conclude that z  0 also. Now go to the
2
2
2
2
2
2
2
constraint, g ( x, y, z )  x  2 y  z  6  0  2 y  0  6  2 y  6  y   3
Case 1)
We get two more points:
1
1
f y  g y  1  ( )  4 y , we have y  . In addition,
2
2
1
f x  g x  z    2 x  z  (2 x)  z  x . Going back to the constraint, we have
2
1
11
11 1 11
x 2  2 y 2  z 2  6  x 2  2( ) 2  x 2  6  x  
, ,
)
. Since x=z, we get two points (
2
2
2 2 2
11 1
11
, ,
).
and ( 
2 2
2
Case 2)

1
:
2
(0, 3, 0), (0, 3,0)
Using
1
1
f y  g y  1  ( )  4 y , we have y   . In addition,
2
2
1
f x  g x  z    2 x  z   (2 x)  z   x . Going back to the constraint, we have
2
1
11
x 2  2 y 2  z 2  6  x 2  2( ) 2  ( x) 2  6  x  
. Since x   z , we get two points
2
2
11 1
11
11 1 11
(
, ,
) and (
, ,
).
2
2 2
2
2
2
Case 3)

1
:
2
Using
We therefore get a total of 6 points that possibly the maximum can be attained..
point
(
f(x,y,z)
11 1 11
, ,
)
2 2 2
11 1
, ,
2 2
11 1
(
, ,
2
2
(
(
11
)
2
11
)
2
11 11 1 13
)(
) 
2
2
2 4
11
11 1 13
)( 
) 
2
2
2 4
11 11 1 9
(
)(
) 
2
2
2 4
(
(
11 1
11
, ,
)
2
2
2
(0, 3, 0)
(
(0, 3, 0)
0 3   3
Thus the maximum value is
11
11 1 9
)( 
) 
2
2
2 4
0 3  3
13
4
6) Compute
  F  n d
for the surface
x 2  y 2  z  4, z  0 , F  x  2 y,2 x,0  .
(it may be
easier not to parametrize the surface)
Solution: The surface is a paraboloid, its projection onto the xy-plane is a circle of radius 2.
  F  n d   
 x  2 y,2 x,0  2 x,2 y,1 
dA  
 2 x,2 y,1  0,0,1 
Computer the integral using polar coordinates:
2
2
0
0
 
2r 2 cos 2   rdrd  ( 
2
2
0
2
cos 2 d )(  r 2 dr ) 
0
 (2 x
2
 4 xy  4 xy)dA
1
8 8
(2 )( ) 
2
3
3
7) Verify the following formula:
\
Solution: This is a solid, so use the triple integral: Rotate the surface about the z-axis. Since the
density is constant, we may assume it is 1.

(x 2  y 2 )dV  
2
0
R2
 
R1
o
h
r 2 rdzdrd   ( 
Iz 
2
0
obtain the formula, we need to next find the mass.
R  R1
1
2
2
I z  (2 )( 2
)( h)  M ( R1  R2 )
4
2
4
4
R2  R1
)( h) .
R1
0
4
2
2
M  V  1(h( R2  R1 )) . Thus
d )( 
R2
h
r 3 dr )(  dz )  (2 )(
4
4
To